Explore topic-wise InterviewSolutions in Current Affairs.

Since, 

Wire is bent in the form of a circle of radius 42 cm. 

∴ The wire length = the circumference of circle = 2π×42 

= 2 × \(\frac{22}{7}\) ×42 

= 44 × 6 

= 264 cm. 

Given that,

The wire again bent in form of a square. 

Let the side length of square is a cm. 

Therefore, 

The perimeter of square = length of the wire. 

∴ 4a = 264 cm 

⇒ a = \(\frac{264}{4}\) = 66 cm. 

Hence,

The area of square = a² = (66)²cm². 

And the area of circle = πr² 

= \(\frac{22}{7}\) × 42 × 42 

= 22 × 6 × 42 cm². 

Now, 

\(\frac{Area\,of\,circle}{Area\,of\,square}\) = \(\frac{22\times 6\times 42}{66\times 66}\) 

= \(\frac{22}{3\times 11}\) = \(\frac{14}{11}\) 

Hence, 

The ratio of areas of the regions enclosed by the circle and the squares is 14:11.

Concept:

Circumference of a circle is given by

S = 2πr

Where r is the radius of a circle

Circumference of a rectangle is given by

P = 2(l + b)

Where l is the length ang b is width of a rectangle

Here A circular ring is cut and bent into the form of a rectangle so perimeter of circle and rectangle are same.

S = 2πr = P = 2(l + b)

2πr = 2(l + b)

Calculation:

Given:

radius r = 42 cm, l : b = 6 : 5

Let l = 6x ,b = 5x

2πr = 2(l + b)

2 π × 42 = 2(6x + 5x)

2 × 22/7 × 42 = 2(11x)

x = 12

So l = 6 × 12 = 72

b = 5 × 12 = 60

The small side of the rectangle is 60 cm.

Let tens digit be x and units digit be y

Then number =10x+y

Since the number is 4 times the sum of digits

Hence 10x+y=4(x+y) i. e. 6x=3y. Or 2x=y……(i)

Also when 27 is added the digits are reversed

Number when digits are reversed =10y+x

Hence 10x+y+27=10y+x

9x+27=9y. Or x+3=y………..(ii)

From (i) and (ii) we get

x+3=2x hence x=3. and y=2x=6

and number =36

Given:

(The place value of 5 in 15201) + (the place value of 6 in 2659) = 7 × ______

Calculations:

The place value of 5 in 15201 = 5 × 1000

⇒ 5000

the place value of 6 in 2659 = 6 × 100

⇒ 600

(The place value of 5 in 15201) + (the place value of 6 in 2659) = 5000 + 600

⇒ 5600

Let  the number in the blank space be 'x'

5600 = 7 × x

⇒ x = 5600 ÷ 7

⇒ x = 800

∴  The number in the blank space is 800

Given:

Number of man days required to complete the work = 56

Calculation:

Let the number of man be m and number of days be d

Number of man days = m × d = 56

Number of man = (m – 1)

⇒ Number of days = (d + 1)

⇒ Number of man days = (m – 1) (d + 1)

⇒ md =(m – 1) (d + 1)

⇒ md = md – d + m – 1

⇒ m – d = 1

putting d = 56/m 

⇒ m – 56/m = 1 

⇒ m² – 56 = m

⇒ m² – m – 56 = 0

⇒ m² – 8m + 7m – 56 = 0

⇒ m(m – 8) + 7(m – 8) = 0

⇒ (m – 8) (m + 7) = 0 

⇒ m = 8, – 7 

m is number of man, so m cannot be negative 

⇒ m = 8 

∴ The Number of workers is 8.

in question we have given

a * b = a+ b̅ 

m = a* b

m * b =?

(a* b) * b

( a+ b̅) * b       // a * b = a+ b̅

(a+ b̅) + b̅      //a + a̅ = a

a + b̅ = a * b = m

Therefore option 2 is correct

Given:

Volume of first solution = 40 liters

Ratio of milk and water in first solution = 3 : 2

Volume of second solution = 30 liters

Ratio of milk and water in second solution = 2 : 1

Calculation:

Volume of milk in first solution = (3/5) × 40 liters

⇒ 24 liters

Volume of water in first solution = (2/5) × 40 liters

⇒ 16 liters

Volume of milk in second solution = (2/3) × 30 liters

⇒ 20 liters

Volume of water in second solution = (1/3) × 30 liters

⇒ 10 liters

Volume of milk in new mixture = Volume of milk in first solution + Volume of milk in second solution

⇒ (24 + 20) liters

⇒ 44 liters

Volume of water in new mixture = Volume of water in first solution + Volume of water in second solution

⇒ (16 + 10) liters

⇒ 26 liters

Ratio of milk and water in mixture = 44/26

⇒ 22/13

⇒ 22 : 13

∴ The ratio of milk and water in the resultant solution is 22 : 13 

Given:

K men can do the work in K days

Formula used:

Time = Total work/Efficiency

Calculation:

Total work = K × K = K²

Time taken by Y to do the work = K²/Y days

∴ The required time taken is K²/Y days

Correct option is: (d) 25 : 4

Analysis:

Y = A B̅ C + A B̅ C̅ 

= AB̅ (C + C̅ )

= AB̅  [Since, C + C̅ = 1]

Correct answer is

(b) Terminating decimal expansion after 2 places of decimals.

Calculation

Numbers are,

a) 1/3

b) 0.3 = 3/10

c) 33/121

d) 0.33 = 33/100

To compare them make the denominator equal, by taking LCM of denominators

⇒ LCM (3,10,121, 100) = 36300

Multiply numerator and denominator by 36300, then the numbers will become

a) (1 × 36300)/(3 × 36300)

⇒ 12100/36300

b) (3 × 36300)/(10 × 36300)

⇒ 10890/36300

c) (33 × 36300)/(121 × 36300)

⇒ 9900/36300

d) (33 × 36300)/(100 × 36300)

⇒ 11979/36300

so, the largest number is 12100/36300

∴ The largest number is 1/3

Correct option is (B) \( \frac{na}{n + 1}\)

∵ ABCD is a parallelogram

∴ AB = CD

∴ \(AE = \frac{AB}n = \frac{CD}n = CF\)   \(\left(\because \frac{AE}{AB} = \frac{CF}{CD} = \frac 1n (given)\right)\)

In \(\triangle CDX\),

\(\triangle CFY \sim\triangle CDX\)

\(\therefore FY \parallel DX\)

⇒ \(\frac{CF}{CD} = \frac{CY}{CX}\)

⇒ \(\frac{CY}{CX} = \frac1n\)

Similarly in \(\triangle ABY\),

\(\frac{AX}{AY} = \frac{AE}{AB} = \frac1n\)

\(\therefore \frac{CY}{CX} = \frac{AX}{AY}\)

Also,

\(\triangle CFY \cong \triangle AXE\)

\(\therefore AX = CY\)

\(\frac{AY}{AX} = n\)

⇒ \(\frac{AY + AX}{AX} = n + 1\)

⇒ \(\frac{AY + CY}{AX} = n + 1\)

⇒ \(AX = \frac{AC}{n + 1}\)

Now,

\(XY = AC - AX - CY\)

\(= AC - 2AX\)

\(= AC - \frac{AC}{n + 1}\)     \(\left(\because AX = \frac{AC}{n + 1}\right)\)

\(= \frac{(n + 1) - 1}{n + 1}AC\)

\(= \frac{n}{n + 1}a\)

Concept

For a number to be divisible by 44, it should be divisible by 4 and 11.

Divisibility by 4 : Last two digits of given number should be divisible by 4

Divisibility by 11 : The sum of the digits at even place is subtracted from the sum of the digits at odd places. If the result is either 0 or divisible by 11, then the original number is divisible by 11

Calculation

Checking divisibility by 4 :

Last two digits of all options are

34, 18, 28 and 08

From the given options, only option 3 and 4 are divisible by 4 since their last two digits are divisible by 4 

Now, checking option 3 and 4 for divisibility by 11 :

3) 86328

(8 + 3 + 8) - (6 + 2) = 19 - 8

⇒ 11, which is divisible by 11

So, option 3 is divisible by 11

4) 86208

(8 + 2 + 8) - (6 + 0) = 12, which is not divisible by 11

Since option 3 is divisible by both 4 and 11, therefore it is divisible by 44.

GIVEN:

A number 6x54y is completely divisible by both 9 and 11.

CALCULATION:

Divisibility by 9 : If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.

Sum of digits = 6 + x + 5 + 4 + y = (15 + x + y)

Possible value of ‘x + y’ = 2 and 12

Divisibility by 11: If the difference between the sum of the odd-numbered digits and the sum of the even-numbered digits, counted from right to left, is divisible by 11, then the number is divisible by 11.

(6 + 5 + y) – (x + 4) = (7 + y – x)

Possible value of ‘y – x’ = 4

Since difference between ‘x’ and ‘y’ is 4 so, its sum cannot be 2.

Hence, sum of ‘x’ and ‘y’ = 12

Left hand limit of function\(f\) at \(x\) = 1 is \(f\)(1 −) = \(​​\lim\limits_{x \to1} f(x)=\lim\limits_{x \to 1}\cfrac{x^2-1}{x-1} \)

\(=\lim\limits_{x \to 1}\cfrac{(x-1)(x+1)}{x-1} \) = \(​​\lim\limits_{x \to1}\,x+ 1\) = 1 + 1 = 2.

Hence, Left hand limit of function\(f\) at \(x\) = 1 is \(f\) (1 ⁻) = 2.

Since, given that the function \(f\)(\(x\)) is continuous at \(x\) = 1.

Therefore, \(f\)(1 ⁻) = \(f\)(1) = \(f\)(1 ⁺), where \(f\)(1 ⁺) is right hand limit of function \(f\)(\(x\)) at \(x\) = 1.

Now, \(f\)(1 ⁻) = \(f\)(1) ⇒ k = 2. ( \(\because\) \(f\)(1) = and \(f\)(1 ⁻) = 2 )

Hence, if k = 2, then the function \(f\) (\(x\)) is continuous at \(x\) = 1.

Given:

x = √(7 + 14√(7 + √(3 + 1)))

Calculations:

√4 = 2, √9 = 3, √49 = 7

x = √(7 + 14√(7 + √(3 + 1)))

⇒ x = √(7 + 14√(7 + √4))

⇒ x = √(7 + 14√(7 + 2))      [√4 = 2]

⇒ x = √(7 + 14√9)

⇒ x = √(7 + 14 × 3)      [√9 = 3]

⇒ x = √(7 + 42)

⇒ x = √49

⇒ x = 7      [√49 = 7]

Now, x - 5 = 7 - 5

⇒ x - 5 = 2

∴ The value of (x - 5) is 2

Given:

Number divided by 30, 36 & 80 and gives a remainder 11.

Concept used:

Using concept of LCM.

Number = Divisor × Dividend + Remainder

Calculation:

The LCM of 30, 36 and 80 is 720.

720 is between 500 and 1000

The required number = 720 + Remainder

⇒ Required number = 720 + 11

⇒ Required number = 731

∴ When 731 is divided by 30, 36 and 80 gives remainder 11.

Given:

Two years back the ratio of ages of Manoj and Sunita = 1 ∶ 2

Three years from now the ratio of their ages will be 3 ∶ 5

Calculations:

Let the present ages of Manoj and Sunita be M and S respectively.

Two years back the ratio of ages of Manoj and Sunita = 1 ∶ 2

⇒ (M – 2)/(S – 2) = 1/2

⇒ 2M – S = 2      ----(1)

Three years from now the ratio of their ages will be 3 ∶ 5

⇒ (M + 3)/(S + 3) = 3/5

⇒ 5M – 3S = -6     ----(2)

Multiply equation (1) with 3

⇒ 3(2M – S) = 2 × 3

⇒ 6M – 3S = 6     ----(3)

Substrate equation (2) from equation (3)

⇒ 6M – 3S – (5M – 3S) = 6 – (-6)

⇒ M = 12

Put the value of M in equation (1)

⇒ 2 × 12 – S = 2

⇒ S = 22

\(\frac{60002-601}6=\frac{59401}6=9900\times6+\frac16\)

Hence, the required remainder is 1.

p - 3i = 3 + 2 (4 + 3i)

⇒ p - 3i = 3 + 8 + 6i

⇒ p - 3i = 11 + 6i

⇒ p = 11 & -q = 6

(By comparing two complex numbers)

⇒ p = 11 & q = 6

a. False. The Environment Permitting Regulations 2010 are concerned with protection of the environment, with the enforcing body being the Environment Agency of England and Wales. 

b. True. 

c. False. The Administration of Radioactive Substance Advisory Committee (ARSAC) gives the certificate for administration of radioactive substances. 

d. True. 

e. True. In nuclear medicine, the dose is kept within the DRLs set by the ARSAC.

a. False. Leakage should be less than 1 mGy. 

b. True. In a general radiology setting, filtration should not be less than 2.5 mm of Al 

c. False. The collimator should be capable of restricting field size down to 5 cm 5 cm.

d. True. 

e. False. The operator should stand at least 2 m from the tube and X-ray source.

Z = 115/0.8 = 143.7 Ω ; cos φ = R/Z = 0.3 

∴ R = 0.3 × 143.7 = 43.1Ω

Now X_L = √(Z² - R²) = √(143.7² - 43.1²) = 137.1 Ω

∴ L = 137.1/2π × 50 = 0.436 H 

Parallel Combination 

Active component of current (drawn by resistance) 

= 0.8 cos φ = 0.8 × 0.3 = 0.24 A; R = 115/0.24 = 479 Ω 

Quadrature component of current (drawn by inductance) 

= 0.8 sin φ = 0.8√(1 - 0.3²) = 0.763A

∴ X_L = 115/0.763 Ω 

∴ L = 115/0.763 × 2π × 50 = 0.48 H

(a) Rock/steel-like would guide the son among betrayals. 

(b) During slack moments, the boy may be betrayed by his trusted friends.

The correct answer is Kashmir Stag.

• Kashmir Stag is the state animal of Jammu & Kashmir.
• It is also known as Hangul.
• Kashmir stag is a Schedule-I animal, according to the Wildlife (Protection) Act, 1972 and classified as Endangered (EN) by the IUCN. 
• The weight of male Kashmir stag is between 150 to 250 kg.
• They are mainly found in Dachigam Sanctuary in Jammu and Kashmir, with some scattered population in other parts of the State and in the Gamagal Sanctuary of Himachal Pradesh.

• Jammu and Kashmir:
  • Capital: Srinagar (summer) and Jammu (winter)
  • Official language: Urdu
  • State Bird: Kashmir flycatcher (as of 2020)
  • State Tree: Chinar
  • State Flower: Lotus
  • Languages Spoken: Urdu, Gujari, Balti, Koshur, Bhadarwahi

The correct answer is 'Uttar Pradesh'.

• Uttar Pradesh is the fourth largest State of India in terms of area.
• The largest state is Rajasthan followed by MP and Maharashtra.
• The smallest state is Goa.

• Areawise rank of UT.

• Population wise top states.

The correct answer is Corrosion.

• There are different types of chemical weathering processes such as solution, hydration, carbonation, oxidation, reduction, and biological.
• Hence corrosion is not a type of chemical weathering.

• Chemical Weathering
  • It is a cluster of weathering processes namely as a solution, hydration, carbonation, oxidation, reduction, and biological.
  • These processes act on rocks to decompose, dissolve or moderate them to a fine clastic state through chemical reactions by oxygen, surface/ soil water, and other acids.
  • Water and air along with heat must be present to speed up all chemical reactions.

• Solution weathering
  • Solution weathering occurs when the solvent is an acidic solution rather than simple water.
• Hydration,
  • Hydration is the chemical addition of water.
  • Minerals take up water and enlarge.
  • This enlargement causes an increase in the volume of the material itself or rock.
  • This process is long and reversible.
  • This may lead to their disintegration of rocks.
• Carbonation,
  • Carbonation is the reaction of bicarbonate and carbonate with minerals.
  • It is a general process that helps the fragmentation of feldspars and carbonate minerals.
  • Carbon dioxide from the soil and atmospheric air is absorbed by water to form carbonic acid that acts like weak acid.
  • These acids act on various surfaces and erode them.
• Oxidation and Reduction,
  • In weathering, oxidation denotes the forming of a mixture of a mineral with oxygen to form hydroxides or oxides.
  • Oxidation happens where there is ready access to the oxygenated water.
  • The minerals commonly involved in this process are manganese, sulfur, iron, etc.
  • In the process of oxidation, rock fragmentation happens due to the disturbance caused by the addition of oxygen.
  • Oxidation and reduction processes are interconnected.​
• Biological activity and weathering
  • Decaying plant and animal matter help in the production of humic, carbonic, and other acids which enhance decay and solubility of some elements.

Statements 'A' and 'B' both are incorrect.

• The weight of water vapour per unit volume of air in a given temperature is called specific humidity.
• The relative humidity is the ratio of water vapour content in the air to its water vapour holding capacity at a given temperature. 
• Hence both A and B statements are incorrect.

• Types of Humidity
  • Absolute Humidity or Specific Humidity
  • Relative Humidity

If you do not obey the traffic rules, you will be prosecuted.

The correct answer is Paradise on Earth.

• Amir Khusrau was a Sufi singer, Poet, and Scholar from India. He said while seeing Kashmir first time that this the "Paradise on Earth".
• He was born in (1253) the Era of Delhi Sultanate at Patiyali (U P). He was an Iconic figure of the cultural heritage of India.
• He mainly wrote his poetry primarily in the Persian Language.
• He was a mystic and a spiritual disciple of Nizamuddin Auliya of Delhi. 
• His first chance to show his inept skills of Poetry was given by Nasir-ud-Din Bughra Khan, the second son of Balban.
• In 1310  Amir Khusrau became a disciple of The Sufi saint of the Chishti Nizamuddin Auliya.
• Amir Khusrau died in October 1325, six months after the death of his master Nizamuddin Auliya.
• He was the first person from India who conferred with the title of " Voice of India " ( Parrot of India ).
• He was regarded as the "Father of Qawwali and " Father of Urdu Literature ".
• Some famous work of Amir Khusrau was Tuhfat us-Sighr (The Gift of Childhood), Tughlaq Nama (Book of the Tughlaqs), Afzal ul-Fawaid (Greatest of Blessings) containing the teachings of Nizamuddin Auliya Etc.
• He was the Inventor of the Sitar in India.

 About Jammu and Kashmir: 

• Jammu and Kashmir is a newly formed Union territory of India and the 8th union territory of India.
• Jammu and Kashmir was the only state in India with a Muslim-majority population.
• Jammu and Kashmir was the only state in India that had special autonomy under Article 370 of the Constitution of India, according to which no law enacted by the Parliament of India this article was abolished in the last year of October 2019.
• Jammu and Kashmir have two capitals state Srinagar and Jammu.

• The current Lieutenant Governor of Jammu and Kashmir is Mr. Manoj Sinha.
• The first Chief Minister of the Jammu and Kashmir was Mehr Chand Mahajan.

(d) All of the above mentioned factors contributed to the downfall of the Mughal Empire. 

(d) He worked for the Hindu-Muslim unity

(a) Large estate owners

Correct option (B) 3’-End

(d) any Court of law within the territory of India

Charan Singh had the shortest tenure as Prime Minister of India.